How do I prove : 4sinAcosAcos2Asin15 all over. :sin2A(tan225-2sinsquared A)is equal to. :root 6 - root 2 over 2
First, recognize that tan225 = 1
4sinAcosAcos2Asin15/[sin2A(tan225-2sinsquared A]
= 2 sin(2A)cos(2A)sin15/[sin2A(tan225-2sin^2A]
= 2 cos(2A)sin15/(1-2sin^2A)
Now recognize that 1 - 2sin^2A = cos(2A), which cancels a term in the numerator, leaving
2 sin15 = 0.5176
That equals [6^1/2) - 2^1/2]/2
(Use the formula sin 15 = sin(45 - 30) to prove that.)
2 sin15 = 2sin45cos30 - 2cos45sin30
= sqrt2*sqrt3/2 - sqrt2*(1/2)
= (sqrt6)/2 - (sqrt2)/2
Note that this is an identity that is true no matter what the value of A is. We got all the A terms to cancel out.
I read that as
(4sinA cosA cos (2A) sin15°)/(sin (2A)(tan225° - 2sin^2 A) = (√6-√2)/2
Using the indentities:
sin 2A = 2sinAcosA and
cos 2A = 1 - 2sin^2 A
LS = (4sinAcosA(1- 2sin^2 A) sin15°)/(2sinAcosA(1-2sin^2 A) )
= 2sin15°
now we know that
sin 15° = sin(45-30)°
= sin45cos30 - cos45sin30
= (√2/2)(√3/2) - (√2/2)(1/2)
= (√6 - √2)/4
then
2sin15° = (√6-√2)/2
= RS
To prove the given equality, we need to simplify the expression on both sides and show that they are equal. Let's start by simplifying the expression on the left-hand side (LHS) and right-hand side (RHS) separately.
LHS:
We are given:
LHS = (4sinAcosAcos2Asin15) / (sin2A(tan225-2sinsquared A))
Step 1: Simplify the numerator:
Using the double-angle formula for cosine, we can express cos2A as 2cos^2(A) - 1:
LHS = (4sinAcosA(2cos^2(A) - 1)sin15) / (sin2A(tan225-2sinsquared A))
Step 2: Simplify the denominator:
Using the double-angle formula for sine, we can express sin2A as 2sin(A)cos(A):
LHS = (4sinAcosA(2cos^2(A) - 1)sin15) / (2sin(A)cos(A)(tan225-2sinsquared A))
Step 3: Simplify the expression further:
We can cancel out the common factor of sin(A) and cancel out 2sin(A)cos(A) from the numerator and denominator:
LHS = (2cos^2(A) - 1)sin15 / (tan225-2sinsquared A)
Now let's simplify the right-hand side (RHS):
RHS = (root 6 - root 2) / 2
To prove the equality, we need to show that LHS = RHS. So, let's compare the simplified LHS and RHS:
LHS = (2cos^2(A) - 1)sin15 / (tan225-2sinsquared A)
RHS = (root 6 - root 2) / 2
The next step is to manipulate the equation further by evaluating the values of sin15 and tan225. However, the values of these trigonometric functions cannot be expressed in simple exact forms. Therefore, to completely prove the equality, you would need to use a calculator or mathematical software to compute sin15 and tan225 and then substitute those values into the equation to compare LHS and RHS.